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shusha [124]
3 years ago
13

Find the midpoint of the segment with the following endpoints.

Mathematics
2 answers:
Brrunno [24]3 years ago
6 0

Answer: x midpoint 6.5 y midpoint -2.5

Step-by-step explanation:

BigorU [14]3 years ago
5 0

Answer:

(6.5,-2.5)

Step-by-step explanation:

midpoint=( X1+X2/2 , Y1+Y2/2)

midpoint= (10+3/2 , 1+-6/2)

midpoint =(6.5,-2.5)

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Find the dimensions of the rectangular corral producing the greatest enclosed area given 320 feet of fencing. (Assume that the l
jenyasd209 [6]

Answer:

The dimensions of the rectangular corral producing the greatest enclosed area  is a square of 80 ft x 80 ft

Step-by-step explanation:

Let

x -----> the length of the rectangular corral in feet

y -----> the width of the rectangular corral in feet

we know that

The area of the rectangular corral is equal to

A=xy -----> equation A

The perimeter of the rectangular corral is equal to

P=2(x+y)

P=320\ ft

so

320=2(x+y)

Simplify

160=(x+y)

y=160-x -----> equation B

substitute equation B in equation A

A=x(160-x)

A=-x^2+160x

This is a vertical parabola open downward

The vertex is a maximum

The x-coordinate of the vertex represent the length for the maximum area

The y-coordinate of the vertex represent the maximum area

Convert the quadratic equation in vertex form

A=-x^2+160x

Factor -1

A=-(x^2-160x)

Complete the square

A=-(x^2-160x+80^2)+80^2

A=-(x^2-160x+80^2)+6,400

Rewrite as perfect squares

A=-(x-80)^2+6,400

The vertex is the point (80,6,400)

so

x=80\ ft

The maximum area is 6,400 ft^2

<em>Find the value of y</em>

y=160-x  ----> y=160-80=80\ ft

therefore

The dimensions of the rectangular corral producing the greatest enclosed area  is a square of 80 ft x 80 ft

4 0
3 years ago
A recycling plant process an average of 1/3 ton of glass each minute. At approximately what rate does the recycling plant proces
guapka [62]
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Please solve it <br>help!!
Mashcka [7]

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What percent of 134 is 66? (To the nearest tenth of a percen t)
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Step-by-step explanation:

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